Answer: B
Given: \( \sec^2\theta = 1 + \tan^2\theta \) i.e., \( m^2 = 1 + n^2 \) \(\therefore m^2 > n^2 \) i.e., \( m > n \)
Given: \( \sec^2\theta = 1 + \tan^2\theta \) i.e., \( m^2 = 1 + n^2 \) \(\therefore m^2 > n^2 \) i.e., \( m > n \)